## skullpatrol 3 years ago Does 0.999...=1?

1. zzr0ck3r

.9999........ is not a number

2. zzr0ck3r

have you taken calculus?

3. zzr0ck3r
4. zzr0ck3r

scroll down to number 4, its actaully an easy proof.

5. zzr0ck3r

you will enjoy calculus:)

6. carson889

Here is an algebraic proof of it: 10x = 9.999999... with x = 0.99999... 10x - x = 9.99999... - 0.999999... 9x = 9 Thus, x = 1

7. zzr0ck3r

This question was asked because I was trying to exaplin that inbetween any two numbers is another number, he then said what about .9999999.... and 1, i then tried to explain to him that .99999999.. was not a finite number, then we had this question posted:)

8. zzr0ck3r

there is no amount of 9's that you can write after .9 that will make it equal to 1, but the limit as the amount of 9's goes to infinity is said to equal 1

9. ByteMe

wait... doesn't 0.9999... = 9/10 + 9/100 + 9/1000 + .... an infinite bounded series?

10. mayankdevnani

The number "0.9999..." can be "expanded" as: 0.9999... = 0.9 + 0.09 + 0.009 + 0.0009 + ...

11. mayankdevnani

12. mayankdevnani

13. zzr0ck3r

lol, read this. then you will see the point of the question http://openstudy.com/study#/updates/50ab1ceee4b06b5e49334d38

14. mayankdevnani

So the formula proves that 0.9999... = 1.

15. mayankdevnani

@skullpatrol ok

16. zzr0ck3r

but .99999...is not a number is the point of this topic

17. mayankdevnani

0.999999..... is a no. we have tp prove that it is equal to 1

18. mayankdevnani

*to

19. zzr0ck3r

its a number if you think about convergence, but you can never write 1 in the form .999999999999, read the last question and you will see the point im trying to make

20. zzr0ck3r

its not a finite number.... it can be looked at as a sequence....at infinity then its a number. I told him that between any two numbers is another number, he then said what about .99999999999..... and 1, this is what followed.

21. zzr0ck3r

Its an extended real number in the since that infinity is an extended real number....

22. mayankdevnani

"But", some say, "there will always be a difference between 0.9999... and 1." Well, sort of. Yes, at any given stop, at any given stage of the expansion, for any given finite number of 9s, there will be a difference between 0.999...9 and 1. That is, if you do the subtraction, 1 – 0.999...9 will not equal zero. But the point of the "dot, dot, dot" is that there is no end; 0.9999... is inifinte. There is no "last" digit. So the "there's always a difference" argument betrays a lack of understanding of the infinite. (That's not a "criticism", per se; infinity is a messy topic.)

23. zzr0ck3r

right and i dont think he has had calculus...

24. zzr0ck3r

this started with density of rational/irational and archimedes principle on the real line

25. zzr0ck3r

the point is, that there is an infinite amount of numbers between any two numbers a,b where a != b

26. mayankdevnani

hahhaha.. we are just fighting you and me are absolutely right...it's a skull problem..right??

27. mayankdevnani

right?? @zzr0ck3r

28. zzr0ck3r

This one might make him want to quit math The infinite amount of numbers between 0,1 is larger than the infinite amount of numbers between 0 and infinity.:)

29. zzr0ck3r

right:)

30. mayankdevnani

:)

31. waterineyes

0.9999 is approximately equal to 1 but it is not exactly equal.. $0.999 \approx 1 \qquad \qquad (0.999 \ne 1)$

32. mayankdevnani

got it @skullpatrol

33. mayankdevnani

this question...lol

34. mayankdevnani
35. Zarkon

can you provide a proof of this ... "This one might make him want to quit math The infinite amount of numbers between 0,1 is larger than the infinite amount of numbers between 0 and infinity.:)"

36. zzr0ck3r

no, my analysis teacher said it two days ago. she said the elemnts between 0 and 1 have more mappings than 0-infinity

37. Zarkon

your teacher is either wrong or you have misquoted her

38. Zarkon

the real numbers between 0 and 1 is the same size as the real numbers from 0 to infinity

39. ParthKohli

$1 - 0.999\cdots = 0.000\cdots = 0$